Either way square

There is an easy way to do it only with a piece of paper:

You write all the squares from 1 to 196 (anymore won't satisify $x\quad +\quad y\quad \le \quad 200$ So you have :

1 4 9 16 25 36 49 64 81 100 121 144 169 196

Then you start with x - y = 1

So (5,4) is a solution, (13,12) , (25,24),... quickly you see that only the odd squares satisfy the condition , that's because they are the solution to the system: $\begin{cases} x & - & y & = & 1 \\ x & + & y & = & square \end{cases}$ , where $x\quad =\frac { square\quad +\quad 1 }{ 2 } \\ y\quad =\frac { square\quad -\quad 1\quad }{ 2 }$

For x - y = 4 , only the even squares satisfy the condition, for the same reason that above.

• For x - y = 1, you get 6 solutions ( from 1, 9, 25,... to 169)
• For x - y = 4, you get 6 solutions ( from 16 , 36, ... to 196)
• For x - y = 9, you get 5 solutions ( from 25 to 196 )
• For x - y = 16, you get 4 solutions ( from 36 to 144)
• For x - y = 25, you get 4 solutions ( from 49 to 169)
• For x - y = 36, you get 3 solutions (etc)
• For x - y = 49, you get 2 solutions
• For x - y = 64, you get 1 solutions
• For x - y = 81, you get 0 solutions

It seems like a lot of work, but you can do it in 2 minutes, you just have to check the higher square so it doesn't surpass x + y <= 200

Adding up all the solutions, you get 31

int ps=0; for(int i=1;i<=100;i++) for(int j=1;j<=100;j++) { if(i!=j&&((Math.sqrt(i-j)-(int)(Math.sqrt(i-j))==0&&Math.sqrt(i+j)-(int)Math.sqrt(i+j)==0))) { ps++;System.out.println(i+" "+j); } } System.out.println("vvalue"+ps);

java code:

public class brilliant201409071548{
    public static void main(String[] args){
        int count = 0;
        for(int y=1;y<=100;y++){
            for(int x=y+1;x<=100;x++){
                if(Math.sqrt(x-y)-((int)(Math.sqrt(x-y)))==0&&Math.sqrt(x+y)-((int)(Math.sqrt(x+y)))==0){
                    count++;
                    System.out.println("("+x+","+y+")");
                }
            }
        }
        System.out.println(count);
    }
}

output:

(5,4)
(10,6)
(17,8)
(26,10)
(13,12)
(37,12)
(50,14)
(20,16)
(65,16)
(82,18)
(29,20)
(25,24)
(40,24)
(53,28)
(34,30)
(68,32)
(45,36)
(85,36)
(41,40)
(58,42)
(52,48)
(73,48)
(90,54)
(65,56)
(61,60)
(80,64)
(74,70)
(97,72)
(89,80)
(85,84)
(100,96)
31
Press any key to continue . . .

Mathematica Code:

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isSq[n_] := Sqrt[n] == Floor[Sqrt[n]]
isSq[0] := False
F[{x_, y_}] := And[isSq[x + y] , isSq[x - y]]

Select[Subsets[Range[100], 2], F]

Outputs:

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5

    {{4, 5}, {6, 10}, {8, 17}, {10, 26}, {12, 13}, {12, 37}, {14, 
  50}, {16, 20}, {16, 65}, {18, 82}, {20, 29}, {24, 25}, {24, 
  40}, {28, 53}, {30, 34}, {32, 68}, {36, 45}, {36, 85}, {40, 
  41}, {42, 58}, {48, 52}, {48, 73}, {54, 90}, {56, 65}, {60, 
  61}, {64, 80}, {70, 74}, {72, 97}, {80, 89}, {84, 85}, {96, 100}}